tìm x
a) \(\left|5x-4\right|=\left|x+2\right|\)
b) \(|\left|2x-1\right|+\frac{1}{2}|\)
c) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
Bài 4 xem lại đề nhé bác
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha
1)\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
2)\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
3)\(\frac{1}{x^2+3x+2}+\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}\)
Tìm x biết
a, \(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
b,\(\left(4x-1\right)^2-\left(2x+3\right)^2+5\left(x+2\right)^2+3\left(x-2\right)\left(x+2\right)=500\)
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(25x^2-10x+1-25x^2+16=7\)
\(17-10x=7\)
\(10x=10\)
\(x=1\)
Bài 3 : Xét dấu biểu thức sau :
1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\)
2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\)
3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\)
4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\)
5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\)
6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\)
7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\)
8 , \(f\left(x\right)=\left(-x^2+x-1\right)\left(6x^2-5x+1\right)\)
9 , \(f\left(x\right)=\frac{x^2-x-2}{-x^2+3x+4}\)
10 , \(f\left(x\right)=\left(x^2-5x+4\right)\left(2-5x+2x^2\right)\)
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
5.
\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)
\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
tìm GTLN
a)\(A=x^2+5y^2+2xy-4x-8y+2015\)
b)\(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
c)\(C=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)
d)\(D=\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)+10\)
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)
\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)
\(B=2x^2+2x+2012^2+2013^2\)
\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)
\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)
\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)
Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)
d) \(^{ }4x\left(2x+3\right)-8x\left(x+4\right)\)
e) \(^{ }2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
f) \(^{ }x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)
\(=8x^2+12x-8x^2-32x\)
=-20x
e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
\(=10x^2+4x+6x^2-2x-9x+3\)
\(=16x^2-7x+3\)
f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)
\(=4x^2+x-4\)
Giải bất phương trình
a) \(\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
b)\(\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
c)\(\left(\frac{3x}{2}-1\right)^2-\frac{x}{2}\left(\frac{5x}{2}-2\right)\le0\)
d) \(\left(2x+1\right)\left(x-3\right)\left(1-5x\right)< 0\)
e)\(2x^2< 16\)
f) \(x^4>8x\)
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
Giải các phương trình sau :
a) \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
b) \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
c) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
d) \(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)
e) \(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
f) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}